Network Network Analyzer 3 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network metrics and an integer K, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints. The target analyzer value is the sum of all metrics greater than or equal to the Kth largest metric.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Analyzer 3"
WHY DOES IT MATTER?
The "K‑largest using a fixed‑size heap" pattern isolates a threshold value without full sorting, which is a common requirement in analytics, leaderboards, and streaming data where only top‑K metrics matter.
OPTIMIZATION CHALLENGE
The key insight is that you only need to track K elements, not the whole dataset. By discarding any element smaller than the current K‑th largest, you reduce both time (log K vs. log N) and space (O(K) vs. O(N)).
REAL-WORLD CONNECTION
Think of a network monitoring dashboard that continuously displays the top K latency spikes. The system keeps a min‑heap of the K highest latencies; older lower values are evicted, ensuring the dashboard always reflects the current critical thresholds.
During an interview, build the heap first, then do a second pass for the sum. If you try to sum while building, you must be careful not to include elements that later get evicted; a clean two‑pass approach avoids that pitfall.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem asks for the sum of all elements that are at least as large as the K‑th largest element in a list. A naïve solution would sort the entire array, locate the K‑th largest value, and then iterate again to accumulate the sum, which costs O(N log N) time. For large N (up to 10^6 or more) this becomes a bottleneck, especially when memory constraints prevent storing a full sorted copy. The optimal paradigm leverages a min‑heap of fixed size K. By maintaining the K largest elements seen so far, the heap’s root always holds the current K‑th largest value. In a single pass we can both keep the heap up‑to‑date (O(log K) per insertion) and accumulate the sum of elements that meet the threshold once the heap is fully built. This yields an O(N log K) time algorithm with O(K) auxiliary space, which is dramatically faster when K ≪ N.
Why this works stems from the heap property: a min‑heap of size K discards any element smaller than its root, guaranteeing that after processing the entire array the root equals the true K‑th largest element. Because we never need the full ordering of the array—only the cutoff value—the heap provides the minimal amount of work to isolate that cutoff. The final summation can be performed in a second linear scan or on‑the‑fly after the heap is finalized, preserving the overall linear‑ith‑log‑K complexity.
Interview Questions on This Problem
Q1How would you find the K‑th largest element in an unsorted array in O(N log K) time?
Maintain a min‑heap of size K. Iterate through the array, pushing each element onto the heap; if the heap size exceeds K, pop the smallest element. After processing all elements, the heap root is the K‑th largest.
Q2Explain why sorting the entire array is sub‑optimal for this problem, and when it might still be acceptable.
Sorting costs O(N log N) which is unnecessary because we only need the K‑th largest value, not the full order. Sorting is acceptable when N is small (e.g., ≤10^4) or when K is close to N, making the heap advantage negligible.
Q3If K can be larger than N, how would you adapt your solution and what would be the expected output?
If K > N, the K‑th largest element is undefined; in this context we treat the cutoff as the smallest element, so the sum becomes the total sum of the array. The algorithm can detect K > N and simply return the sum of all elements.
Examples
Input
[4, 5, 8, 2], 2
Output
13
Explanation: Step-by-step: with input [4, 5, 8, 2] and K = 2, we first find the 2nd largest metric which is 5. Then we sum all metrics greater than or equal to 5, which are 5 and 8. So the target analyzer value is 5 + 8 = 13.
Input
[1, 2, 3, 4, 5], 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we first find the 3rd largest metric which is 3. Then we sum all metrics greater than or equal to 3, which are 3, 4, and 5. So the target analyzer value is 3 + 4 + 5 = 12.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of size K to find the K‑th largest in O(N log K), then sum qualifying elements in a second linear scan.
Brute Force Approach
Sort the entire array, pick the K‑th largest element, then iterate once to sum all elements ≥ that value.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let kthLargest = nums[k - 1];
return nums.filter(num => num >= kthLargest).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int kthLargest = nums[k - 1];
int sum = 0;
for (int num : nums) {
if (num >= kthLargest) {
sum += num;
}
}
return sum;
}
};import java.util.Arrays;
public class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int kthLargest = nums[nums.length - k];
int sum = 0;
for (int num : nums) {
if (num >= kthLargest) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
kth_largest = nums[k - 1]
return sum(num for num in nums if num >= kth_largest)function solution(nums, k) {
nums.sort((a, b) => b - a);
let kthLargest = nums[k - 1];
return nums.filter(num => num >= kthLargest).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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